Given points, p1, p2, p3, p4 ... pn representing a polyline; how do I calculate the geometry (specially vertices - for drawing in graphics applications) assuming a given thickness d. (As in the diagram below)
Specifically need to find a, b, and c. Note that line p1 a is perpendicular to line p1 p2. Line p2 b is not perpendicular to line p1 p2 and also not to line p2 p3.
b is the intersection of the line b c and line a b which are parallel to line p2 p3 and line p1 p2 respectively.
d' is not important, but may prove useful.
Also since all points are represented on the screen as 2 tuples - (x, y); it would be helpful to know a, b and c in the 2 tuple form.
I am thoroughly blanking on how to do this. Any pointers at all will be helpful.
Consider the 2D vector $\vec P_{12}$ from $P_1$ to $P_2$. You normalize it to unit length by dividing by its components by its length
$$L_{12}=\sqrt{(P_{2x}-P_{1x})^2+(P_{2y}-P_{1y})^2}$$ and get $\vec p_{12}=(p_{12x},p_{12y})$.
The vector $\vec n_{12}=(-p_{12y},p_{12x})$ is perpendicular to it and also has unit length.
Then $$a=P_1+d\,\vec n_{12}.$$ Similarly, $$c=P_3+d\,\vec n_{23}.$$
$b$ is a little harder. It lies on the bissectrix of $P_{12}$ and $P_{23}$, with a direction vector obtained by the sum $\vec n_{12}+\vec n_{23}$, which you need to normalize to get $\vec n_{123}$.
The distance $d'$ is the same as $d$ with a correction factor obtained by trigonometry from the dot product of $\vec n_{12}$ and $\vec n_{123}$.
Then,
$$b=P_2+d\frac{\vec n_{123}}{\vec n_{12}\cdot \vec n_{123}}.$$